What is the integral of 6xsqrt(x+2) ?

The integral of 6xsqrt(x+2) is 6(2/5 (x+2)^{5/2}-4/3 (x+2)^{3/2})+C

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integral of 6xsqrt(x+2)

\int 6 x x + 2 d x

6 (\frac{2}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}}) + C

Solution steps

\int 6 x x + 2 d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x x + 2 d x

Apply u-substitution

: \int (u - 2) u d u

= 6 \cdot \int (u - 2) u d u

Expand

(u - 2) u : u^{\frac{3}{2}} - 2 u

= 6 \cdot \int u^{\frac{3}{2}} - 2 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int u^{\frac{3}{2}} d u - \int 2 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 2 u d u = \frac{4}{3} u^{\frac{3}{2}}

= 6 (\frac{2}{5} u^{\frac{5}{2}} - \frac{4}{3} u^{\frac{3}{2}})

Substitute back

u = x + 2

= 6 (\frac{2}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}})

Add a constant to the solution

= 6 (\frac{2}{5} (x + 2)^{\frac{5}{2}} - \frac{4}{3} (x + 2)^{\frac{3}{2}}) + C

\int 8 \frac{x}{4} + 1 d x

\frac{d}{d x} (8 x^{2} + x)

n = 1 \sum \infty (\frac{1}{2})^{n - 1}

\int \frac{t ^{2}}{t ^{4} + 1} d t

\frac{\partial}{\partial y} (6 e^{yz} cos (x yz))

What is the integral of 6xsqrt(x+2) ?
The integral of 6xsqrt(x+2) is 6(2/5 (x+2)^{5/2}-4/3 (x+2)^{3/2})+C